Question

Solve: $2{\log }_4(4-x)=4-{\log }_2(2+x)$

Answer 1

$2{\log }_4(4-x)=4-{\log }_2(2+x)$

$\Rightarrow 2{\log }_{2^2}(4-x)=4-{\log }_2(x+2)$

$\Rightarrow {\log }_2(4-x)+{\log }_2(x+2)=4$

$\Rightarrow {\log }_2(4-x)(x+2)=4$

$\Rightarrow (4-x)(2+x)=2^4=16$

$\Rightarrow 8+4x-x^2-2x-16=0$

$\Rightarrow x^2-2x+8=0$

$\Rightarrow x^2-2x+1-1+8=0$

$\Rightarrow {(x-1)}^2=-7$

$\Rightarrow x-1=\pm \sqrt{-7}$

$\therefore x=1\pm i\sqrt{7}$

Thus, there is no solution.